question form

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phy 201

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Question: 1) When masses of 35, 70 and 105 grams are hung from a certain rubber band its respective lengths are observed to be 39, 46, and 53 cm. 
What are the x and y components of the tension of a rubber band of length 50.6 cm if the x component of its length if 14.02232?
What horizontal force, when added to this force will result in a total force of magnitude 150 grams?

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Your self-critique:

Given solution:

STUDENT SOLUTION WITH INSTRUCTOR COMMENTS INDICATED BY **
I am assuming that you plug in what you know into the equation of the line. 
Y = .2x + 32
** If y represents the mass and x the length that approach would give you the right mass--good thinking, though the equation would be more like y = 5 x - 150. Your equation looks reasonably good if y is length and x is supported mass. However you do need to use the force, which will be the force exerted by gravity on the mass. Force and mass are not the same thing.
In general you will use the graph to determine the force corresponding to a given length. If there is significant curvature to the graph the linear model won?t work very well. In that case you could just draw a smooth curve to fit the data and read your forces from the curve. **
When the length is 50.6 cm the corresponding weight is 93 grams and when the length is 14.02232 cm the weight is ?89.9 g, but I don?t think weight can be negative????
** Plug the rubber band length into the force relationship or read the force from your graph. If your graph is of supported mass vs. length you need to find the force. 
In this case you have a length of 50.6 cm, which corresponds to a supported mass of about 93 grams. This corresponds to a force of about .093 kg * 9.8 m/s^2 = .91 Newtons.
The direction of the rubber band is arctan(y/x). Using the Pythagorean Theorem with length 50.6 cm and x component 14 cm we get y component about 48.6 cm, so the direction of the rubber band is arctan(48.6 / 14) = 74 deg. The angle at which the tension acts is parallel to the rubber band, so the tension also acts at 74 deg.
The x and y components of the force are therefore 
Fx = .91 N * cos(74 deg) = .25 N and Fy = .91 N * sin(74 deg) = .87 N.
If a horizontal force Fhoriz is added then the x component becomes Fx = .25 N + Fhoriz and the magnitude of the resulting force is 
| F | = sqrt(Fx^2 + Fy^2) = sqrt( (.25 N + Fhoriz)^2 + (.87 N)^2). 
If this force is 150 grams (or, converting this to a force, 1.47 N) then we have 
sqrt( (.25 N + Fhoriz)^2 + (.87 N)^2) = 1.47 N. Squaring both sides we get
.0625 N^2 + .5 N * Fhoriz + Fhoriz^2 + .76 N^2 = 2.16 N^2. 
This is quadratic in Fhorz and rearranges to Fhoriz^2 + .5 N * Fhoriz ? 2.1 N^2 = 0. Using the quadratic formula we get 
Fhoriz = (-.5 N +- sqrt( (.5 N)^2 ? 4 * 1 * (-2.1 N^2) ) / (2 * 1) = (-.5 N +- 2.8 N) / 2, giving us two forces:
Fhoriz = (-.5 N + 2.8 N ) / 2 = 1.15 N and 
Fhoriz = (-.5 N ? 2.8 N) / 2 = -1.65 N.**

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I see a copy of the problem, but I don't see your question.

One problem is that this does not seem to be in text format.

This problem comes directly from the Introductory Problem Sets, so you should also reference that source.

The point is that 35, 70 and 105 gram masses have weights of about .35, .69 and 1.03 Newtons, so you can from this information get a graph of force vs. length. From that graph you can find the tension of a 50 cm rubber band, which gives you a force to use in solving the problem.

From the 50.6 cm length and the 14 cm x component of the length you can find the angle of the rubber band, and hence the force, with the positive x axis. So you can find the components of the force.

Using the components of the force you can answer the question.

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